Inner and Outer Product of the Wavefunctions

[Dario56]

Inner product is a generalization of the dot product on spaces other than Euclidean and for vectors it is defined in the same way as the dot product. If we have two vectors $v$ and $w$, than their inner product is: $$\langle v|w\rangle = v_1w_1 + v_2w_2 + ...+v_nw_n $$

where $v_1,w_1, v_2,w_2...v_n,w_n$ are the corresponding components of both vectors in particular basis set.

We know that wavefunctions are vectors in the Hilbert space. This comes from the linear properties of quantum systems (Schrödinger equation). Namely, we can express every wavefunction as the linear combination of particular complete and orthonormal basis set: $$ |\Psi\rangle = \sum_k c_k|\psi_k\rangle$$If we have two wavefunctions, $\Psi_1$ and $\Psi_2$ than their inner product is defined as: $$ \langle \Psi_1|\Psi_2\rangle = \sum_{i,j} c_{1,i}^*c_{2,j} \langle \psi_i|\psi_j\rangle $$

where we expanded wavefunctions in some basis set.

To calculate the inner product, we need to define what it means to take the inner product between functions generally and we need to see how it applies to the basis functions. Inner product between two functions in general $\psi_i$ and $\psi_j$ is defined differently compared to vectors although definition is similar: $$ \langle\psi_i|\psi_j\rangle = \int\psi_i^* \psi_j \,dx $$

Idea is that we multiply both functions for every value of $x$ and than sum over all possible values of $x$. Since $x$ is continuous, we integrate rather than sum.

If the basis set is orthonormal than inner product of two basis functions in the Hilbert space must be equal to Kronecker delta, $\delta_{ij}$ in the similar way as it is the case for orthonormal vectors in Euclidean space (dot product): $$ \langle\psi_i|\psi_j\rangle = \int\psi_i^* \psi_j \,dx = \delta_{ij}$$

$$ \langle \Psi_1|\Psi_2\rangle = \sum_{i,j} c_{1,i}^*c_{2,j} \delta_{ij} = \sum_i c_{1,i}c_{2,i}$$

Outer product of the wavefunction $\Psi$ with itself can be defined as: $$ |\Psi\rangle \langle \Psi| = \sum_{i.j} c_i |\psi_i\rangle c_j^* \langle \psi_j| = \sum_{i,j} c_ic_j^* |\psi_i\rangle \langle \psi_j| $$

where we expressed the wavefunction in the basis set.

I am not sure how to calculate the outer product between the basis functions? For the inner product, I know how is it defined generally and I know it must be equal to Kronecker delta in the case of basis functions.

I also know that sum of the outer products of basis functions should be equal to the identity, but I don't know how to define and calculate it in the case of individual basis functions.
functions.

 

[DrClaude]

We know that wavefunctions are vectors in the Hilbert space. This comes from the linear properties of quantum systems (Schrödinger equation). Namely, we can express every wavefunction as the linear combination of particular complete and orthonormal basis set: $$ \Psi = \sum_k c_k\psi_k $$

We can express the last equation in the vector form (Dirac notation) where we write it as the inner product: $$ \Psi = \langle c|\psi\rangle = c^\dagger \psi $$

That's not correct. You are confusing vectors and scalars.

The equivalent of $$ \Psi = \sum_k c_k\psi_k $$ in Dirac notation is $$ \ket{\Psi} = \sum_k c_k \ket{\psi_k}$$

If we have two wavefunctions, $\Psi_1$ and $\Psi_2$ than their inner product is defined as: $$ \langle \Psi_1|\Psi_2\rangle = \sum_{i,j} c_i^*c_j \langle \psi_i|\psi_j\rangle $$
where we expanded wavefunctions in some basis set.

Your notation is sloppy here and causes problem later. $\ket{\Psi_1}$ and $\ket{\Psi_2}$ being distinct, you cannot use the same coefficients $c$ for both. So instead of $c_i^*c_j$, you should have. something like $c_{1,i}^*c_{2,j}$.

$$ \langle \Psi_1|\Psi_2\rangle = \sum_{i,j} c_i^*c_j \delta_{ij} = \sum_i |c_i|^2 $$

$$ \langle \Psi_1|\Psi_2\rangle = \sum_{i,j} c_{1,i}^*c_{2,j} \delta_{ij} = \sum_i c_{1,i}^*c_{2,i} $$

Outer product of the wavefunction $\Psi$ with itself can be defined as: $$ |\Psi\rangle \langle \Psi| = \sum_k c_k |\psi_k\rangle c_k^* \langle \psi_k| = \sum_k |c_k|^2 |\psi_k\rangle \langle \psi_k| $$

where we expressed the wavefunction in the basis set.

No, it is a double sum
$$ |\Psi\rangle \langle \Psi| = \sum_{i,j} c_i |\psi_i\rangle c_j^* \langle \psi_j| $$
The outer product is an operator (equivalent to a matrix, not a vector).

 

[Dario56]

That's not correct. You are confusing vectors and scalars.

The equivalent of $$ \Psi = \sum_k c_k\psi_k $$ in Dirac notation is $$ \ket{\Psi} = \sum_k c_k \ket{\psi_k}$$Your notation is sloppy here and causes problem later. $\ket{\Psi_1}$ and $\ket{\Psi_2}$ being distinct, you cannot use the same coefficients $c$ for both. So instead of $c_i^*c_j$, you should have. something like $c_{1,i}^*c_{2,j}$.

$$ \langle \Psi_1|\Psi_2\rangle = \sum_{i,j} c_{1,i}^*c_{2,j} \delta_{ij} = \sum_i c_{1,i}^*c_{2,i} $$

No, it is a double sum
$$ |\Psi\rangle \langle \Psi| = \sum_{i,j} c_i |\psi_i\rangle c_j^* \langle \psi_j| $$
The outer product is an operator (equivalent to a matrix, not a vector).

That's not correct. You are confusing vectors and scalars.

The equivalent of $$ \Psi = \sum_k c_k\psi_k $$ in Dirac notation is $$ \ket{\Psi} = \sum_k c_k \ket{\psi_k}$$

Yes, I made a mistake here. I corrected this.

Your notation is sloppy here and causes problem later. $\ket{\Psi_1}$ and $\ket{\Psi_2}$ being distinct, you cannot use the same coefficients $c$ for both. So instead of $c_i^*c_j$, you should have. something like $c_{1,i}^*c_{2,j}$.

Yup, this is indeed not correct. Will correct it. It is easy to make a mistake in the notation when you write in very general terms.

No, it is a double sum
$$ |\Psi\rangle \langle \Psi| = \sum_{i,j} c_i |\psi_i\rangle c_j^* \langle \psi_j| $$
The outer product is an operator (equivalent to a matrix, not a vector).

Double sum is generally correct. However, since I have the same wavefunction $\Psi$ is it reasonable to expand it in the same basis set which eliminates the need to write the double sum?

 

[DrClaude]

Double sum is generally correct. However, since I have the same wavefunction $\Psi$ is it reasonable to expand it in the same basis set which eliminates the need to write the double sum?

It is expanded in the same basis here, namely $\ket{\psi}$. There is no further simplification possible.

 

[Dario56]

It is expanded in the same basis here, namely $\ket{\psi}$. There is no further simplification possible.

Oh, yes. That is correct. Now, when all mistakes are corrected, we can get to my original question which is: How to calculate outer product of two basis functions? $$ |\psi_i\rangle\langle\psi_j| $$

As I said in the main post, inner product of the mentioned functions is defined as: $$ \langle \psi_i |\psi_j \rangle = \int \psi_i^* \psi_j \,dx$$

I can't really find what is the general definition for the outer product of two functions, though.

 

[DrClaude]

It's an operator. It transforms a state into another state. How it will act specifically depends on the representation.

 

[topsquark]

...I can't really find what is the general definition for the outer product of two functions, though.

If you are looking for something like integral form of finding the inner product, you aren't going to get anywhere. Like DrClaude said, the outer product is going to be a matrix, or operator.For example, in the bra-ket notation, let's look at the spin 1/2 system. Let's write + and - state in the "usual" basis:
$|+> = \left ( \begin{matrix} 1 \\ 0 \end{matrix} \right )$

$|-> = \left ( \begin{matrix} 0 \\ 1 \end{matrix} \right )$

Then we can get things like the outer product
$|+><-| = \left ( \begin{matrix} 1 \\ 0 \end{matrix} \right ) \left ( \begin{matrix} 0 & 1 \end{matrix} \right ) = \left ( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right )$

The exact form of the outer product depends on the basis we use to define the wavefunctions.

-Dan

 

[Dario56]

If you are looking for something like integral form of finding the inner product, you aren't going to get anywhere. Like DrClaude said, the outer product is going to be a matrix, or operator.For example, in the bra-ket notation, let's look at the spin 1/2 system. Let's write + and - state in the "usual" basis:
$|+> = \left ( \begin{matrix} 1 \\ 0 \end{matrix} \right )$

$|-> = \left ( \begin{matrix} 0 \\ 1 \end{matrix} \right )$

Then we can get things like the outer product
$|+><-| = \left ( \begin{matrix} 1 \\ 0 \end{matrix} \right ) \left ( \begin{matrix} 0 & 1 \end{matrix} \right ) = \left ( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right )$

The exact form of the outer product depends on the basis we use to define the wavefunctions.

-Dan

Yes, so outer product can't be defined for functions in the same way as the inner product can. We can only define it, if we represent quantum states with vectors in certain basis like you did. Since quantum states are vectors in the Hilbert space, this isn't a problem.

 

[DrClaude]

Of course, you can write out what the projector looks like when applied to a wave function:
$$
\ket{\phi} \braket{\phi | \Psi} \doteq \phi \int \phi^* \Psi dx
$$